Lesson 18 (10/05/26): Method of Undetermined Coefficients (3.5)

Lesson 18 (10/05/26): Method of Undetermined Coefficients (3.5)

Review: Mechanical Vibrations

\[ x'' + \frac{c}{m} x' + \omega_0^2 x = F(t), \quad \omega_0 = \sqrt{\frac{k}{m}} \]

① No friction: \( c = 0 \), No external force: \( F(t) = 0 \rightsquigarrow \text{Undamped} \)

\[ x'' + \omega_0^2 x = 0 \]

General sol:

\[ x(t) = c_1 \cos \omega_0 t + c_2 \sin \omega_0 t \]

rewrite using trigonometry

\[ = A \cos(\omega_0 t - \phi) \]

\( A = \text{Amplitude} \)

\( \phi = \text{phase difference} \)

\( T = \frac{2\pi}{\omega_0} \quad \text{period.} \)

2. \( c \neq 0 \) (there is friction), \( F(t) = 0 \) No external force — "Damped"

\[ x'' + \frac{c}{m} x' + \omega_0^2 x = 0 \]

Ch. equation: \[ r^2 + \frac{c}{m} r + \omega_0^2 = 0 \]

\[ r = \frac{-c \pm \sqrt{c^2 - 4m^2 \omega_0^2}}{2m} \]

i) \( c^2 - 4m^2 \omega_0^2 > 0 \) — "Over" Damped

has two \( - \)ve roots \( r_1, r_2 \)

Sol: \[ c_1 e^{r_1 t} + c_2 e^{r_2 t} \]

ii) \( c^2 - 4m^2 \omega_0^2 = 0 \) — "Critically Damped"

has one root \( r = -\frac{c}{2m} \)

Sol: \[ c_1 e^{r t} + c_2 t e^{r t} \]

iii) \( c^2 - 4m^2 \omega_0^2 < 0 \) — "Under" Damped

has ch roots \( \alpha \pm i\beta \), \( \alpha = -\frac{c}{2m} < 0 \)

Sol: \[ e^{\alpha t} [c_1 \cos \beta t + c_2 \sin \beta t] \]

\[ x'' + \frac{c}{m} x' + \omega_0^2 x = F(t) \]

\( F(t) \neq 0 \rightsquigarrow \) Non-Homogeneous, Linear \( 2^{\text{nd}} \) order ODE

We will learn 2 methods

1. Method of Undetermined Coefficients (Today)

   when const. coeff.
   +
   Right hand side is simple:

   • "polynomial"
   • exponential
   • trig.

2. Variation of Parameters (After Fall break)

eg: find a solution to \( y'' - 3y' + 2y = 2x + 5 \)

if \( y \) is a solution
it is a function whose linear combination & \( y, y', y'' \) is a \( 1^{\text{st}} \) order polynomial

Guess: Look for a \( 1^{\text{st}} \) order polynomial as a solution

Suppose \( y = Ax + B \) is a solution
\( y' = A, \quad y'' = 0 \)

want \( y'' - 3y' + 2y = 2x + 5 \)

\[ 0 - 3A + 2(Ax + B) = 2x + 5 \] \[ (2A)x + (-3A + 2B) = 2x + 5 \]

Compare coeff: \( 2A = 2 \Rightarrow A = 1 \), \( -3A + 2B = 5 \Rightarrow B = 4 \)

\( y = x + 4 \) is a solution.

eg: \( y'' - 6y' - 7y = 18e^{-2x} \)

Look for solution of the form \( y = Ae^{-2x} \)

\[ y' = -2Ae^{-2x}, \quad y'' = 4Ae^{-2x} \]

want: \( y'' - 6y' - 7y = 18e^{-2x} \)

\[ 4Ae^{-2x} + 12Ae^{-2x} - 7Ae^{-2x} = 18e^{-2x} \]

\[ 9Ae^{-2x} = 18e^{-2x} \implies 9A = 18 \implies A = 2 \]

\( y = 2e^{-2x} \) is one solution to

\[ y'' - 6y' - 7y = 18e^{-2x} \]

\[ y'' - 6y' - 7y = 18e^{-2x} \]

\( y_p = 2e^{-2x} \) is one solution

1. Suppose \( y_c \) is a solution to \( y'' - 6y' - 7y = 0 \). Is \( y_c + y_p \) a solution to \( y'' - 6y' - 7y = 18e^{-2x} \)? YES

2. How do we find all the solution?

Verification for 1:

\( 1 \Rightarrow y = y_c + y_p, \quad y' = y_c' + y_p', \quad y'' = y_c'' + y_p'' \)

\[ \begin{aligned} y'' - 6y' - 7y &= (y_c'' + y_p'') - 6(y_c' + y_p') - 7(y_c + y_p) \\ &= \underbrace{(y_c'' - 6y_c' - 7y_c)}_{\to 0} + \underbrace{(y_p'' - 6y_p' - 7y_p)}_{\to 18e^{-2x}} \\ &= 18e^{-2x} \end{aligned} \]

Answer for 2:

Add all solutions to the homogeneous to one solution of non-homogeneous.

\[ a_n(x) y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_2(x) y'' + a_1(x) y' + a_0(x) y = F(x) \]

1. Find complementary solution \( y_c(x) \) which is the general solution to

   \[ a_n(x) y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_2(x) y'' + a_1(x) y' + a_0(x) y = 0 \]

2. Find one solution \( y_p(x) \) called the particular solution (using undetermined coeff. or variation of parameters) to

   \[ a_n(x) y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_2(x) y'' + a_1(x) y' + a_0(x) y = F(x) \]

\( y = y_c + y_p \) is the General Solution.

\[ y'' - 6y' - 7y = 18 e^{-2x} \]

Find \( \underline{y_c} \): ch. equation \( r^2 - 6r - 7 = 0 \Rightarrow r = -1, \, r = 7 \)

\[ y_c = C_1 e^{-x} + C_2 e^{7x} \]

is general solution to

\[ y'' - 6y' - 7y = 0 \]

Already found using undetermined coeff

\( y_p = 2e^{-2x} \) is solution to

\[ y'' - 6y' - 7y = 18 e^{-2x} \]

\( y = C_1 e^{-x} + C_2 e^{7x} + 2e^{-2x} \) is the General Solution.

Example: \( y'' + y = \sin 2x \)

Complementary solution:

\[ y'' + y = 0 \]

Ch. equation: \( r^2 + 1 = 0 \implies r = \pm i \)

\[ y_c = c_1 \sin x + c_2 \cos x \]

Particular solution: using undetermined coeff.

Look for

\[ y_p = A \cos 2x + B \sin 2x \] \[ y_p' = -2A \sin 2x + 2B \cos 2x \] \[ y_p'' = -4A \cos 2x - 4B \sin 2x \]

want: \( y_p'' + y_p = \sin 2x \)

\[ -4A \cos 2x - 4B \sin 2x + A \cos 2x + B \sin 2x = \sin 2x \] \[ -3A \cos 2x - 3B \sin 2x = \sin 2x \]

Compare coeff: \( A = 0, \quad B = -\frac{1}{3} \implies y_p = -\frac{1}{3} \sin 2x \)

Gen. Sol:

\[ y = y_c + y_p = c_1 \sin x + c_2 \cos x - \frac{1}{3} \sin 2x \]

Method of Undetermined Coefficient

\[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_2 y'' + a_1 y' + a_0 y = F(x) \]

Constant coeff, Linear Non-homogeneous
and \( F(x) \) is simple "poly", "Trig", "Exponential"

Look for \( y_p \) that looks like \( F(x) \)

eg.

\[ F(x) = 2x + 5 \rightsquigarrow y_p = Ax + B \] \[ F(x) = 18e^{-2x} \rightsquigarrow y_p = Ae^{-2x} \] \[ F(x) = \sin 2x \rightsquigarrow y_p = A \sin 2x + B \cos 2x \] \[ F(x) = 9x^2 + 5 + 21e^{7x} \rightsquigarrow y_p = Ax^2 + Bx + C + De^{7x} \]

\[ y'' + y = \sin x \]

What goes wrong?